CBSE Class 10 Maths HOTs Arithmetic Progressions Set I

Very Short Answer Type Questions

Question. For what value of \( k \) will \( k + 9, 2k - 1 \) and \( 2k + 7 \) are the consecutive terms of an A.P.
Answer: If \( x, y \) and \( z \) are consecutive terms of an A.P. then we have
\( y - x = z - y \)
Thus if \( k + 9, 2k - 1, \) and \( 2k + 7 \) are consecutive terms of an A.P. then we have
\( (2k - 1) - (k + 9) = (2k + 7) - (2k - 1) \)
\( 2k - 1 - k - 9 = 2k + 7 - 2k + 1 \)
\( k - 10 = 8 \)
\( k = 10 + 8 = 18 \)

Question. What is the common difference of an A.P. in which \( a_{21} - a_7 = 84 \)?
Answer: Let the first term of an A.P. be \( a \) and common difference be \( d \).
\( a_{21} - a_7 = 84 \)
\( [a + (21 - 1)d] - [a + (7 - 1)d] = 84 \)
\( a + 20d - a - 6d = 84 \)
\( 14d = 84 \)
\( d = 6 \)

Question. In the A.P. \( 2, x, 26 \) find the value of \( x \).
Answer: If \( x, y, \) and \( z \) are in AP then we have
\( y - x = z - y \)
Since \( 2, x \) and \( 26 \) are in A.P. we have
\( x - 2 = 26 - x \)
\( 2x = 26 + 2 \)
\( x = \frac{28}{2} = 14 \)

Question. For what value of \( k \); \( k + 2, 4k - 6, 3k - 2 \) are three consecutive terms of an A.P.
Answer: If \( x, y \) and \( z \) are three consecutive terms of an A.P. then we have
\( y - x = z - y \)
Since \( k + 2, 4k - 6 \) and \( 3k - 2 \) are three consecutive terms of an AP, we obtain
\( (4k - 6) - (k + 2) = (3k - 2) - (4k - 6) \)
\( 4k - 6 - k - 2 = 3k - 2 - 4k + 6 \)
\( 3k - 8 = -k + 4 \)
\( 4k = 4 + 8 \)
\( k = \frac{12}{4} = 3 \)

Question. If \( 18, a, b, -3 \) are in AP, then find \( a + b \).
Answer: If \( 18, a, b, -3 \) are in AP, then,
\( a - 18 = -3 - b \)
\( a + b = -3 + 18 \)
\( a + b = 15 \)

Question. Find the sum of first five multiples of 2.
Answer: Let the first term be \( a \), common difference be \( d \), \( n^{th} \) term be \( a_n \) and sum of \( n^{th} \) term be \( S_n \)
Here, \( a = 2, d = 2, n = 5 \)
\( S_n = \frac{n}{2}[2a + (n - 1)d] \)
\( S_5 = \frac{5}{2}[2 \times 2 + (5 - 1)2] \)
\( = \frac{5}{2}[4 + 4 \times 2] = \frac{5}{2}[4 + 8] \)
\( = \frac{5}{2} \times 12 = 5 \times 6 = 30 \)

Question. Find the sum of first 16 terms of the A.P. 10, 6, 2, .....
Answer: Let the first term be \( a \), common difference be \( d \), \( n^{th} \) term be \( a_n \) and sum of \( n \) term be \( S_n \)
Here, \( a = 10, d = 6 - 10 = -4, n = 16 \)
\( S_n = \frac{n}{2}[2a + (n - 1)d] \)
\( S_{16} = \frac{16}{2}[2 \times 10 + (16 - 1)(-4)] \)
\( = 8[20 + 15 \times (-4)] \)
\( = 8[20 - 60] \)
\( = 8 \times (-40) \)
\( = -320 \)

Question. What is the sum of five positive integer divisible by 6.
Answer: Let the first term be \( a \), common difference be \( d \), \( n^{th} \) term be \( a_n \) and sum of \( n^{th} \) term be \( S_n \)
Here, \( a = 6, d = 6, n = 5 \)
\( S_n = \frac{n}{2}[2a + (n - 1)d] \)
\( S_5 = \frac{5}{2}[2 \times 6 + (5 - 1)(6)] \)
\( = \frac{5}{2}[12 + 4 \times 6] \)
\( = \frac{5}{2}[12 + 24] = \frac{5}{2}[36] \)
\( = 5 \times 18 = 90 \)

Question. Find the common difference of the A.P. \( \frac{1}{3q}, \frac{1 - 6q}{3q}, \frac{1 - 12q}{3q}, \dots \)
Answer: Let common difference be \( d \) then we have
\( d = \frac{1 - 6q}{3q} - \frac{1}{3q} \)
\( = \frac{1 - 6q - 1}{3q} = \frac{-6q}{3q} = -2 \)

Question. Find the first four terms of an A.P. whose first term is \( 3x + y \) and common difference is \( x - y \).
Answer: Let the first term of an A.P. be \( a \) and common difference be \( d \).
Now \( a_1 = 3x + y \)
\( a_2 = a_1 + d = 3x + y + x - y = 4x \)
\( a_3 = a_2 + d = 4x + x - y = 5x - y \)
\( a_4 = a_3 + d = 5x - y + x - y = 6x - 2y \)
So, the four terms are \( 3x + y, 4x, 5x - y \) and \( 6x - 2y \).

Question. Find the \( 37^{th} \) term of the A.P. \( \sqrt{x}, 3\sqrt{x}, 5\sqrt{x} \)
Answer: Let the \( n^{th} \) term of an A.P. be \( a_n \) and common difference be \( d \).
Here, \( a_1 = \sqrt{x} \)
\( a_2 = 3\sqrt{x} \)
\( d = a_2 - a_1 = 3\sqrt{x} - \sqrt{x} = 2\sqrt{x} \)
\( a_n = a + (n - 1)d \)
\( a_{37} = \sqrt{x} + (37 - 1)2\sqrt{x} \)
\( = \sqrt{x} + 36 \times 2\sqrt{x} \)
\( = 73\sqrt{x} \)

Question. For an A.P., if \( a_{25} - a_{20} = 45 \), then find the value of \( d \).
Answer: Let the first term of an A.P. be \( a \) and common difference be \( d \).
Now \( a_{25} - a_{20} = \{a + (25 - 1)d\} - \{a + (20 - 1)d\} \)
\( 45 = a + 24d - a - 19d \)
\( 45 = 5d \)
\( d = \frac{45}{5} = 9 \)

Question. Find, 100 is a term of the A.P. 25, 28, 31,...... or not.
Answer: Let the first term of an A.P. be \( a \), common difference be \( d \) and number of terms be \( n \).
Let \( a_n = 100 \)
Here \( a = 25, d = 28 - 25 = 31 - 28 = 3 \)
Now \( a_n = a + (n - 1)d \),
\( 100 = 25 + (n - 1) \times 3 \)
\( 100 - 25 = 75 = (n - 1) \times 3 \)
\( 25 = n - 1 \)
\( n = 26 \)
Hence, 100 is a term of the given A.P.

Question. Is 184 a term of the sequence 3, 7, 11,.......?
Answer: Let the first term of an A.P. be \( a \), common difference be \( d \) and number of terms be \( n \).
Let \( a_n = 184 \)
Here, \( a = 3, d = 7 - 3 = 11 - 7 = 4 \)
Now \( a_n = a + (n - 1)d \),
\( 184 = 3 + (n - 1)4 \)
\( \frac{181}{4} = n - 1 \)
\( 45.25 = n - 1 \)
\( 46.25 = n \)
Since 46.25 is not an whole number, thus 184 is not a term of given A.P.

Question. Find the \( 7^{th} \) term from the end of A.P. 7, 10, 13, .... 184.
Answer: Let us write A.P. in reverse order i.e., 184,..... 13, 10, 7
Let the first term of an A.P. be \( a \) and common difference be \( d \).
Now \( d = 7 - 10 = -3 \)
\( a = 184, n = 7 \)
\( 7^{th} \) term from the end,
\( a_7 = a + 6d \)
\( a_7 = 184 + 6(-3) \)
\( = 184 - 18 = 166 \).
Hence, 166 is the \( 7^{th} \) term from the end.

Question. Which term of an A.P. 150, 147, 144, ..... is its first negative term?
Answer: Let the first term of an A.P. be \( a \), common difference be \( d \) and \( n^{th} \) term be \( a_n \).
For first negative term \( a_n < 0 \)
\( a + (n - 1)d < 0 \)
\( 150 + (n - 1)(-3) < 0 \)
\( 150 - 3n + 3 < 0 \)
\( -3n < -153 \)
\( n > 51 \)
Therefore, the first negative term is \( 52^{nd} \) term.

Question. In an A.P. the sum of first \( n \) terms is \( \frac{3n^2}{2} + \frac{13n}{2} \). Find the \( 25^{\text{th}} \) term.
Answer: We have \( S_n = \frac{3n^2 + 13n}{2} \)
\( a_n = S_n - S_{n-1} \)
\(a_{25} = S_{25} - S_{24}\)
\(= \frac{3(25)^2 + 13(25)}{2} - \frac{3(24)^2 + 13(24)}{2}\)
\(= \frac{1}{2} \{3(25^2 - 24^2) + 13(25 - 24)\}\)
\(= \frac{1}{2} (3 \times 49 + 13) = 80\)

Short Answer Type Questions

Question. A thief runs with a uniform speed of 100 m/minute. After one minute a policeman runs after, the thief to catch him. He goes with a speed of 100 m/minute in the first minute and increased his speed by 10 m/minute every succeeding minute. After how many minutes the policeman will catch the thief. 
Answer: Let total time to catch the thief be \( n \) minutes
Total distance covered by thief \( = 100n \)
Total distance covered by policeman \( = 100 + 110 + 120 + \dots + (n - 1) \text{ terms} \)
\( 100n = \frac{n - 1}{2}[200 + (n - 2)10] \)
\( n^2 - 3n - 18 = 0 \)
\( (n - 6)(n + 3) = 0 \)
\( n = 6 \)
Policeman takes 5 minutes to catch the thief.

Question. If \( S_n \) denotes the sum of first \( n \) terms of an A.P., Prove that, \( S_{30} = 3(S_{20} - S_{10}) \).
Answer: Let the first term be \( a \), and common difference be \( d \).
Now \( S_{30} = \frac{30}{2}(2a + 29d) \) ...(1)
\( = 15(2a + 29d) \)
\( 3(S_{20} - S_{10}) = 3[ \frac{20}{2}(2a + 19d) - \frac{10}{2}(2a + 9d) ] \)
\( = 3[10(2a + 19d) - 5(2a + 9d)] \)
\( = 3[20a + 190d - 10a - 45d] \)
\( = 3[10a + 145d] \)
\( = 15[2a + 29d] \) ...(2)
Hence \( S_{30} = 3(S_{20} - S_{10}) \)

Question. The sum of first 20 terms of an A.P. is 400 and sum of first 40 terms is 1600. Find the sum of its first 10 terms. 
Answer: Let the first term be \( a \), common difference be \( d \), \( n^{th} \) term be \( a_n \) and sum of \( n \) term be \( S_n \)
We know \( S_n = \frac{n}{2}[2a + (n - 1)d] \)
Now \( S_{20} = \frac{20}{2}(2a + 19d) \)
\( 400 = 10(2a + 19d) \)
\( 2a + 19d = 40 \) (1)
Also, \( S_{40} = \frac{40}{2}(2a + 39d) \)
or, \( 1600 = 20(2a + 39d) \)
or, \( 2a + 39d = 80 \) (2)
Solving (1) and (2), we get \( a = 1 \) and \( d = 2 \).
Now \( S_{10} = \frac{10}{2}[2 \times 1 + (10 - 1)2] \)
\( = 5[2 + 9 \times 2] \)
\( = 5[2 + 18] \)
\( = 5 \times 20 = 100 \)

Question. Find \( (4 - \frac{1}{n}) + (7 - \frac{2}{n}) + (10 - \frac{3}{n}) + \dots \) upto \( n \) terms.
Answer: Let sum of \( n \) term be \( S_n \), then we have
\( s_n = (4 - \frac{1}{n}) + (7 - \frac{2}{n}) + (10 - \frac{3}{n}) + \dots \text{ upto } n \text{ terms.} \)
\( = (4 + 7 + 10 + \dots \text{ upto } n \text{ terms}) - (\frac{1}{n} + \frac{2}{n} + \frac{3}{n} + \dots + \frac{n}{n}) \)
\( = (4 + 7 + 10 + \dots + n \text{ terms}) - \frac{1}{n}(1 + 2 + 3 + \dots + n) \)
\( = \frac{n}{2}[2 \times 4 + (n - 1)(3)] - \frac{1}{n} \times \frac{n}{2}[2 \times 1 + (n - 1)(1)] \)
\( = \frac{n}{2}[8 + 3n - 3] - \frac{1}{2}[2 + n - 1] \)
\( = \frac{n}{2}(3n + 5) - \frac{1}{2}(n + 1) \)
\( = \frac{3n^2 + 5n - n - 1}{2} \)
\( = \frac{3n^2 + 4n - 1}{2} \)

Question. Find the \( 60^{th} \) term of the A.P. 8, 10, 12, ...., if it has a total of 60 terms and hence find the sum of its last 10 terms. 
Answer: Let the first term be \( a \), common difference be \( d \), \( n^{th} \) term be \( a_n \) and sum of \( n \) term be \( S_n \)
We have \( a = 8, d = 10 - 8 = 2 \)
\( a_n = a + (n - 1)d \)
Now \( a_{60} = 8 + (60 - 1)2 = 8 + 59 \times 2 = 126 \)
and \( a_{51} = 8 + 50 \times 2 = 8 + 100 = 108 \)
Sum of last 10 terms,
\( S_{51-60} = \frac{n}{2}(a_{51} + a_{60}) \)
\( = \frac{10}{2}(108 + 126) \)
\( = 5 \times 234 = 1170 \)
Hence sum of last 10 terms is 1170.

Question. An arithmetic progression 5, 12, 19, .... has 50 terms. Find its last term. Hence find the sum of its last 15 terms.
Answer: Let the first term be \( a \), common difference be \( d \), \( n^{th} \) term be \( a_n \) and sum of \( n \) term be \( S_n \)
We have \( a = 5, d = 12 - 5 = 7 \) and \( n = 50 \)
\( a_{50} = 5 + (50 - 1)7 = 5 + 49 \times 7 = 348 \)
Also the first term of the A.P. of last 15 terms be \( a_{36} \)
\( a_{36} = 5 + 35 \times 7 = 5 + 245 = 250 \)
Now, sum of last 15 terms
\( S_{36-50} = \frac{15}{2}[a_{36} + a_{50}] \)
\( = \frac{15}{2}[250 + 348] \)
\( = \frac{15}{2} \times 598 = 4485 \)
Hence, sum of last 15 terms is 4485.

Question. Find the middle term of the sequence formed by all three-digit numbers which leave a remainder 3 when divided by 4. Also find the sum of all numbers on both sides of the middle terms separately.
Answer: Let the first term be \(a\), common difference be \(d\), \(n^{th}\) term be \(a_n\) and sum of \(n\) term be \(S_n\).
The three digit numbers which leaves 3 as remainder when divided by 4 are: 103, 107, 111, ..., 999.
Now, the first number \(a = 103\), last number \(a_n = 999\) and common difference \(d = 4\).
Let the number of terms in this sequence be \(n\).
\(a_n = a + (n - 1)d\)
\(999 = 103 + (n - 1)4\)
\(896 = (n - 1)4\)
\((n - 1) = \frac{896}{4} = 224\)
\(n = 224 + 1 = 225\)
Middle term = \(\frac{225 + 1}{2} = 113^{th}\) term
\(a_{113} = 103 + 112 \times 4 = 551\)
and \(a_{112} = 551 - 4 = 547\)
Sum of Ist 112 terms:
\(S_{112} = \frac{112}{2} (a + a_{112})\)
\(= 56(103 + 547)\)
\(= 56 \times 650 = 36400\)
and \(a_{114} = 551 + 4 = 555\)
The sum of last 112 terms:
\(S_{last 112} = \frac{112}{2} (a_{114} + a_{225})\)
\(= 56(555 + 999)\)
\(= 56 \times 1554 = 87024\)

Question. Find the middle term of the sequence formed all numbers between 9 and 95, which leave a remainder 1 when divided by 3. Also find the sum of the numbers on both sides of the middle term separately.
Answer: The sequence is 10, 13, ..., 94
Let the first term be \(a\), common difference be \(d\), \(n^{th}\) term be \(a_n\) and sum of \(n\) term be \(S_n\).
\(94 = 10 + (n - 1)3\)
\(84 = (n - 1)3\)
\(n = \frac{84}{3} + 1 = 29\)
Therefore \(\frac{29 + 1}{2} = 15^{th}\) term is the middle term.
Middle term \(a_{15} = a + (15 - 1)d\)
\(= 10 + 14 \times 3 = 52\)
\(a_{16} = 52 + 3 = 55\)
Sum of first 14 terms,
\(s_{14} = \frac{14}{2} [2 \times 10 + (14 - 1) \times 3]\)
\(= 7[20 + 13 \times 3] = 413\)
\(S_n = \frac{n}{2} [2a + (n - 1)d]\)
Sum of the last 14 terms,
\(= \frac{14}{2} [2a_{16} + (14 - 1)d]\)
\(= \frac{14}{2} [2 \times 55 + (14 - 1) \times 3]\)
\(= 7[110 + 13 \times 3]\)
\(= 1043\)

Question. Find the middle term of the sequence formed by all three-digit numbers which leave a remainder 5 when divided by 7. Also find the sum of all numbers on both sides of the middle term separately.
Answer: Let the first term be \(a\), common difference be \(d\), \(n^{th}\) term be \(a_n\) and sum of \(n\) term be \(S_n\).
The sequence is 103, 110, ..., 999
Here \(a = 103\), \(d = 7\), \(a_n = 999\)
\(a_n = a + (n - 1)d\)
\(999 = 103 + (n - 1) \times 7\)
\(n = \frac{999 - 103}{7} + 1 = 129\)
Therefore \(\frac{129 + 1}{2} = 65^{th}\) term is the middle term.
Middle term \(a_{65} = 103 + (64 \times 7) = 551\)
\(a_{66} = 551 + 7 = 558\)
Sum of first 64 terms,
\(S_{64} = \frac{64}{2} [2a + (64 - 1)d]\)
\(= 32[2 \times 103 + 63 \times 7]\)
\(= 32[206 + 441] = 20704\)
Sum of last 64 terms
\(S_{66-129} = \frac{64}{2} (558 + 999)\)
\(= 32 \times 1557\)
\(= 49824\)

Question. If the sum of first \(n\) term of an A.P. is given by \(S_n = 3n^2 + 4n\). Determine the A.P. and the \(n^{th}\) term.
Answer: Let the first term be \(a\), common difference be \(d\), \(n^{th}\) term be \(a_n\) and sum of \(n\) term be \(S_n\).
We have \(S_n = 3n^2 + 4n\).
\(a_1 = S_1 = 3(1)^2 + 4(1) = 7\)
\(a_1 + a_2 = S_2 = 3(2)^2 + 4(2)\)
\(= 12 + 8 = 20\)
\(a_2 = S_2 - S_1 = 20 - 7 = 13\)
\(a + d = 13\)
or, \(7 + d = 13\)
Thus \(d = 13 - 7 = 6\)
Hence AP is 7, 13, 19, .......
Now, \(a_n = a + (n - 1)d\)
\(= 7 + (n - 1)(6)\)
\(= 7 + 6n - 6\)
\(= 6n + 1\)
\(a_n = 6n + 1\)

Question. The sum of the \(3^{rd}\) and \(7^{th}\) terms of an A.P. is 6 and their product is 8. Find the sum of first 20 terms of the A.P.
Answer: Let the first term be \(a\), common difference be \(d\), \(n^{th}\) term be \(a_n\) and sum of \(n\) term be \(S_n\).
We have \(a_3 + a_7 = 6\)
\(a + 2d + a + 6d = 6\)
\(a + 4d = 3\) --- (1)
and \(a_3 \times a_7 = 8\)
\((a + 2d)(a + 6d) = 8\) --- (2)
Substituting the value \(a = (3 - 4d)\) in (2) we get
\((3 - 4d + 2d)(3 - 4d + 6d) = 8\)
or, \((3 + 2d)(3 - 2d) = 8\)
or, \(9 - 4d^2 = 8\)
\(4d^2 = 1 \implies d^2 = \frac{1}{4} \implies d = \pm \frac{1}{2}\)
CASE 1 : Substituting \(d = \frac{1}{2}\) in equation (1), \(a = 1\).
\(S_{20} = \frac{20}{2} [2a + (n - 1)d]\)
\(= \frac{20}{2} [2 + \frac{19}{2}] = 115\)
Thus \(d = \frac{1}{2}\), \(a = 1\) and \(S_{20} = 115\)
CASE 2 : Substituting \(d = -\frac{1}{2}\) in equation (1), \(a = 5\)
\(S_{20} = \frac{20}{2} [2 \times 5 + 19 \times (-\frac{1}{2})]\)
\(= 10 [10 - \frac{19}{2}] = 15\)
Thus \(d = -\frac{1}{2}\), \(a = 5\) and \(S_{20} = 15\)

Question. A sum of Rs. 280 is to be used towards four prizes. If each prize after the first is Rs. 20 less than its preceding prize, find the value of each of the prizes.
Answer: Let \(I^{st}\) prize be Rs. \(x\), then series of prize is
\(x, x-20, x-40, x-60, \dots\)
Here series is AP and \(a = x, d = -20, S_n = 280, n = 4\)
\(S_n = \frac{n}{2} [2a + (n - 1)d]\)
\(280 = \frac{4}{2} [2x + 3(-20)]\)
\(280 = 2[2x - 60]\)
\(140 = 2x - 60\)
\(x = \frac{140 + 40}{2} = 100\)
Thus prizes are Rs. 100, Rs. 80, Rs. 60, Rs. 40.

Question. In a garden bed, there are 23 rose plants in the first row, 21 are in the \(2^{nd}\), 19 in \(3^{rd}\) row and so on. There are 5 plants in the last row. How many rows are there of rose plants? also find the total number of rose plants in the garden.
Answer: The number of rose plants in the \(1^{st}, 2^{nd}, \dots\) are 23, 21, 19, ..., 5.
Let the first term be \(a\), common difference be \(d\), \(n^{th}\) term be \(a_n\) and sum of \(n\) term be \(S_n\).
Here \(a = 23, d = -2, a_n = 5\)
\(a_n = a + (n - 1)d\)
\(5 = 23 + (n - 1)(-2)\)
\(n = 10\)
Total number of rose plants in the flower bed,
\(S_n = \frac{n}{2} [2a + (n - 1)d]\)
\(S_{10} = 5(46 - 18) = 140\)

Question. A sum of Rs. 1890 is to be used to given seven cash prizes to students of a school for their overall academic performance. If each prize is Rs. 50 less than its preceding prize, find the value of each of the prizes.
Answer: Let \(I^{st}\) prize be Rs. \(x\), then series of prize is
\(x, x-50, x-100, x-150, \dots\)
Here series is AP and
\(a = x, d = -50, S_n = 1890, n = 7\)
\(S_n = \frac{n}{2} [2a + (n - 1)d]\)
\(1890 = \frac{7}{2} [2x + (-50) \times 6]\)
\(270 = x + (-50) \times 3 = x - 150\)
\(x = 270 + 150 = 420\)
The prizes are Rs. 420, Rs. 370, Rs. 320, Rs. 270, Rs. 220, Rs. 170, Rs. 120.

Question. If the sum of first \(m\) terms of an A.P. is same as the sum of its first \(n\) terms (\(m \neq n\)), show that the sum of its first (\(m + n\)) terms is zero.
Answer: Let the first term be \(a\), common difference be \(d\), \(n^{th}\) term be \(a_n\), and sum of \(n\) term be \(S_n\).
Now \(S_m = S_n\)
\(\frac{m}{2} [2a + (m - 1)d] = \frac{n}{2} [2a + (n - 1)d]\)
\(2a(m - n) + \{(m^2 - n^2) - (m - n)\}d = 0\)
\(2a(m - n) + \{(m - n)(m + n) - (m - n)\}d = 0\)
\((m - n)[2a + (m + n - 1)d] = 0\)
\(2a + (m + n - 1)d = 0\) [Since \(m - n \neq 0\)]
\(S_{m+n} = \frac{m + n}{2} [2a + (m + n - 1)d]\)
\(= \frac{m + n}{2} [0] = 0\)

Question. A man repays a loan of Rs. 3250 by paying Rs. 20 in the first month and then increases the payment by Rs. 15 every month. How long will it take him to clear the loan?
Answer: Let the first term be \( a \), common difference be \( d \), \( n^{th} \) term be \( a_n \) and sum of \( n \) term be \( S_n \)
Here \( a = 20, d = 15 \)
Now \( S_n = 3250 \)
\( S_n = \frac{n}{2}[2a + (n - 1)d] \)
\( 3250 = \frac{n}{2}[2a + (n - 1) \times 15] \)
\( 3250 \times 2 = n[40 + 15n - 15] \)
\( 6500 = n[25 + 15n] \)
\( 1300 = n[5 + 3n] \)
\( 3n^2 + 65n - 60n - 1300 = 0 \)
\( n(3n + 65) - 20(3n + 65) = 0 \)
\( (n - 20)(3n + 65) = 0 \)
Since \( n = -65/3 \), is not possible, \( n = 20 \)
Man will repay loan in 20 months.

Question. If \( 1 + 4 + 7 + 10 + \dots + x = 287 \), Find the value of \( x \).
Answer: Let the first term be \( a \), common difference be \( d \), \( n^{th} \) term be \( a_n \) and sum of \( n \) term be \( S_n \)
We have \( a = 1, d = 3 \)
\( S_n = \frac{n}{2}[2a + (n - 1)d] \)
\( \frac{n}{2}[2 \times 1 + (n - 1)3] = 287 \)
\( \frac{n}{2}[2 + (3n - 3)] = 287 \)
\( 3n^2 - n = 574 \)
\( 3n^2 - n - 574 = 0 \)
\( 3n^2 - 42n + 41n - 574 = 0 \)
\( 3n(n - 14) + 41(n - 14) = 0 \)
\( (n - 14)(3n + 41) = 0 \)
Since negative value is not possible, \( n = 14 \)
\( a_{14} = a + (n - 1)d \)
\( = 1 + 13 \times 3 = 40 \)

Question. Find the sum of first 24 terms of an A.P. whose \( n^{th} \) term given by \( a_n = 3 + 2n \).
Answer: Let the first term be \( a \), common difference be \( d \), \( n^{th} \) term be \( a_n \) and sum of \( n \) term be \( S_n \)
We have \( a_n = 3 + 2n \)
\( a_1 = 3 + 2 \times 1 = 5 \)
\( a_2 = 3 + 2 \times 2 = 7 \)
\( a_3 = 3 + 2 \times 3 = 9 \)
Thus the series is 5, 7, 9, ..... in which
\( a = 5 \) and \( d = 2 \)
Now \( S_n = \frac{n}{2}[2a + (n - 1)d] \)
\( S_{24} = \frac{24}{2}(2 \times 5 + 23 \times 2) \)
\( = 12 \times 56 \)
Hence, \( S_{24} = 672 \)